Question

Two particles each with charge +2.77 µC are located on the x axis. One is at x = 1.00 m, and the other is at x = −1.00 m.

(a) Determine the electric potential on the y axis at y = 0.520 m.

Your response is off by a multiple of ten. kV

(b) Calculate the change in electric potential energy of the system as a third charged particle of -2.98 µC is brought from infinitely far away to a position on the y axis at y = 0.520 m.
J

Answer 1

a)

OA = OB = 1 m

OP = 0.520 m

r = AP = BP = distance of each charge from point P on Y-axis

using pythagorean theorem

r = sqrt(OA² + OP²) = sqrt((1)² + (0.520)²) = 1.13 m

q = charge at each A and B = 2.77 x 10^-6 C

Total electric potential at P is given as

V_p = V_a + V_b = k q/r + kq/r = 2 kq/r = 2 (9 x 10⁹) (2.77 x 10^-6)/(1.13)

V_p = 4.4 x 10⁴ Volts

b)

Q = charge brought from infinity = - 2.98 x 10^-6 C

change in electric potential energy is given as

U = Q V_p = (- 2.98 x 10^-6 ) (4.4 x 10⁴) = - 0.13 J